By rotation, we assume that the maximal eigenvalue of \(b^{ij}\)  at \(t\)  is attained at point \(x_{t}\)  in the direction
of the unit vector \(e_{1} \in T_{x_{t}} \mathbb {S}^{n-1}\) . We also choose orthonormal vector field such that \(b^{ij}\)  is diagonal.
By the evolution equation of \(b^{ij}\)  in Lemma REF , we get
\(\begin{split}& \partial _{t}\frac{b^{11}}{h} - N\eta (t)\sigma _{k}^{pq}\nabla _{pq}\frac{b^{11}}{h} \\& = \frac{2}{h}N\eta (t)\sigma _{k}^{pq} \nabla _{p}\frac{b^{11}}{h}\nabla _{q}h + \frac{N}{h^{2}}\eta (t) b^{11}\sigma _{k}^{pq}\nabla _{pq}h\\&- (k+1) \frac{N}{h}\eta (t)\sigma _{k}(b^{11})^{2} + \frac{N}{h}\eta (t)\sigma _{k}^{pq}\delta _{pq}b^{11} \\& - \frac{N}{h}\eta (t)(b^{11})^{2}(\sigma _{k}^{pq,mn} + 2\sigma _{k}^{pm}b^{nq})\nabla _{1}b_{pq}\nabla _{1}b_{mn}\\& -\frac{\eta (t)}{h}(b^{11})^{2}( \nabla _{11}N\sigma _{k} + 2\nabla _{1}\sigma _{k}\nabla _{1}N) - \frac{b^{11}}{h^{2}}N\eta (t)\sigma _{k}+ \frac{2b^{11}}{h}\\&= \frac{2}{h}N\eta (t)\sigma _{k}^{pq} \nabla _{p}\frac{b^{11}}{h}\nabla _{q}h - (k+1) \frac{N}{h}\eta (t)\sigma _{k}(b^{11})^{2} \\& - \frac{N}{h}\eta (t)(b^{11})^{2}(\sigma _{k}^{pq,mn} + 2\sigma _{k}^{pm}b^{nq})\nabla _{1}b_{pq}\nabla _{1}b_{mn}\\& -\frac{\eta (t)}{h}(b^{11})^{2}(\nabla _{11}N\sigma _{k} + 2\nabla _{1}\sigma _{k}\nabla _{1}N)+ (k-1) \frac{b^{11}}{h^{2}}N\eta (t)\sigma _{k}+ \frac{2b^{11}}{h}.\end{split}\) 
According to inverse concavity of \((\sigma _{k})^{\frac{1}{k}}\) , we obtain by [1]} or [2]}
\((\sigma _{k}^{pq,mn} + 2\sigma _{k}^{pm}b^{nq})\nabla _{1}b_{pq}\nabla _{1}b_{mn} \ge \frac{k+1}{k}\frac{(\nabla _{1}\sigma _{k})^{2}}{\sigma _{k}}.\) 
On the other hand, by Schwartz inequality, the following inequality holds
\(2|\nabla _{1}\sigma _{k}\nabla _{1}N| \le \frac{k+1}{k}\frac{N(\nabla _{1}\sigma _{k})^{2}}{\sigma _{k}} + \frac{k}{k+1}\frac{\sigma _{k}(\nabla _{1}N)^{2}}{N}.\) 
Hence, we have at \((x_{t}, t)\)
\(\partial _{t}\frac{b^{11}}{h}\le -\frac{(b^{11})^{2}}{h}\sigma _{k}\eta (t)\left[\nabla _{11}N -\frac{k}{k+1}\frac{(\nabla _{1}N)^{2}}{N} + (k+1)N + (1-k)\frac{Nb_{11}}{h}\right]+ \frac{2b^{11}}{h}.\) 
Let \(\tau \)  be the arc-length of the great circle passing through \(x_{t}\)  with the unit tangent vector \(e_{1}\) .
Notice that
\(\nabla _{11}N -\frac{k}{k+1}\frac{(\nabla _{1}N)^{2}}{N} + (k+1)N = (k+1)N^{\frac{k}{k+1}}\left(N^{\frac{1}{k+1}} + (N^{\frac{1}{k+1}})_{\tau \tau }\right).\) 
Since
\(\begin{split}N_{\tau } & = (f^{-1})_{\tau } \varphi h + f^{-1} \varphi h_{\tau }\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)\\N_{\tau \tau } & = (f^{-1})_{\tau \tau } \varphi h + 2(f^{-1})_{\tau } \varphi h_{\tau }\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right) + f^{-1} \varphi ^{\prime }h_{\tau }^{2}\left(1 + \frac{\varphi ^{\prime }h}{\varphi } \right)\\&+ f^{-1} \varphi h_{\tau \tau }\left(1 + \frac{\varphi ^{\prime }h}{\varphi } \right) + f^{-1} \varphi h_{\tau }^{2}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)^{\prime },\end{split}\) 
here \(f^{-1}\)  is \(\dfrac{1}{f}\) .
We have by direct computations
\(\begin{split}& 1+ N^{-\frac{1}{k+1}}\left(N^{\frac{1}{k+1}}\right)_{\tau \tau } \\&= 1 + \frac{1}{k+1}N^{-1}N_{\tau \tau } - \frac{k}{(k+1)^{2}}N^{-2}N_{\tau }^{2}\\&= 1 + \frac{1}{k+1}f(f^{-1})_{\tau \tau } + \frac{2f}{(k+1)h}(f^{-1})_{\tau }h_{\tau }\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right) \\& + \frac{\varphi ^{\prime }}{(k+1)\varphi h}h_{\tau }^{2}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right) + \frac{h_{\tau \tau }}{(k+1)h}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)\\& + \frac{h_{\tau }^{2}}{(k+1)h}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)^{\prime } - \frac{k}{(k+1)^{2}}f^{2}(f^{-1})_{\tau }^{2}\\& - \frac{2kf}{(k+1)^{2}h}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)(f^{-1})_{\tau }h_{\tau } - \frac{kh_{\tau }^{2}}{(k+1)^{2}h^{2}}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)^{2}\\&= 1 + \frac{1}{k+1}f(f^{-1})_{\tau \tau } + \frac{2f}{(k+1)^{2}h}(f^{-1})_{\tau }h_{\tau }\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right) \\& + \frac{h_{\tau \tau }}{(k+1)h}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right) + \frac{h_{\tau }^{2}}{(k+1)h}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)^{\prime } - \frac{k}{(k+1)^{2}}f^{2}(f^{-1})_{\tau }^{2}\\& +\frac{h_{\tau }^{2}}{(k+1)^{2}h^{2}}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)\left(\frac{\varphi ^{\prime }h}{\varphi } - k\right)\\&= \frac{1 + \frac{\varphi ^{\prime }h}{\varphi } }{k+1}\frac{h_{\tau \tau } + h}{h} + \frac{h_{\tau }^{2}}{(k+1)h}\left(1+ \frac{\varphi ^{\prime }h}{\varphi } \right)^{\prime } \\& -\frac{1+\frac{\varphi ^{\prime }h}{\varphi } }{h(k+1)^{2}}f\left[h_{\tau }\left(\frac{k - \frac{\varphi ^{\prime }h}{\varphi } }{fh}\right)^\frac{1}{2} - (f^{-1})_{\tau }\left(\frac{hf}{k - \frac{\varphi ^{\prime }h}{\varphi } }\right)^\frac{1}{2} \right]^{2}\\& + \frac{1}{k+1}\left[\left(k - \frac{\varphi ^{\prime }h}{\varphi } \right) - (f^{-1})_{\tau }^{2}f^{2}\left(\frac{k}{k+1} + \frac{1}{k+1} \frac{1+\frac{\varphi ^{\prime }h}{\varphi } }{\frac{\varphi ^{\prime }h}{\varphi } - k}\right)+ (f^{-1})_{\tau \tau }f\right]\\& \ge \frac{1 + \frac{\varphi ^{\prime }h}{\varphi } }{k+1}\frac{h_{\tau \tau } + h}{h} \\& + \frac{1}{k+1}\left[\left(k - \frac{\varphi ^{\prime }h}{\varphi } \right) - (f^{-1})_{\tau }^{2}f^{2}\left(\frac{k}{k+1} + \frac{1}{k+1} \frac{1+\frac{\varphi ^{\prime }h}{\varphi }}{\frac{\varphi ^{\prime }h}{\varphi } - k}\right)+ (f^{-1})_{\tau \tau }f\right],\end{split}\) 
where in the last inequality we use the conditions \(\dfrac{\varphi ^{\prime }h}{\varphi } \le -1\)  and \((\dfrac{\varphi ^{\prime }h}{\varphi } )^{\prime } \ge 0.\) 
Since \((k + 1)f^{-\frac{1}{k + a}}e_{ij} + (k + a)(f^{-\frac{1}{k + a}})_{ij}\)  is positive definite and \(-a \le \dfrac{\varphi ^{\prime }h}{\varphi } \le -1, \)  thus we can estimate
\(\begin{split}& \left(k - \frac{\varphi ^{\prime }h}{\varphi } \right) - (f^{-1})_{\tau }^{2}f^{2} \frac{k - \frac{\varphi ^{\prime }h}{\varphi } - 1}{k - \frac{\varphi ^{\prime }h}{\varphi } }+ (f^{-1})_{\tau \tau }f \\& \ge k + 1 - (f^{-1})_{\tau }^{2}f^{2} \frac{k + a - 1}{k + a }+ (f^{-1})_{\tau \tau }f \\& = k + 1 + (k + a)f^{\frac{1}{k + a}}(f^{-\frac{1}{k + a}})_{\tau \tau }\\& = f^{\frac{1}{k + a}}\left[(k + 1)f^{-\frac{1}{k + a}} + (k + a)(f^{-\frac{1}{k + a}})_{\tau \tau }\right]\\& \ge c_{f},\end{split}\) 
where \(c_{f}\)  is a positive constant depending on \(f\)  and the minimal eigenvalue of \((k + 1)f^{-\frac{1}{k + a}}e_{ij} + (k + a)(f^{-\frac{1}{k + a}})_{ij}\) .
Now we can derive
\(\partial _{t} \frac{b^{11}}{h} \le -\left(\frac{b^{11}}{h}\right)^{2}N\sigma _{k}\eta (t)( c_{f}h + (2 - a - k)b_{11}) + \frac{2b^{11}}{h}.\) 
By the uniform bounds on \(h\) , \(f\) , \(\eta \)  and \(\sigma _{k}\) , we conclude
\(\partial _{t}\frac{b^{11}}{h} \le -c_{1}\left(\frac{b^{11}}{h}\right)^{2} + c_{2}\frac{b^{11}}{h}.\) 
Here \(c_{1}\)  and \(c_{2}\)  are positive constants independent of \(t\) .
The maximum principle then gives the upper bound of \(b^{11} \)  and the result follows.
